What is the vertices y=(x+1)(x+2) ?

The vertices y=(x+1)(x+2) is Minimum (-3/2 ,-1/4)

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vertices y=(x+1)(x+2)

vertices

y = (x + 1) (x + 2)

Minimum (- \frac{3}{2}, - \frac{1}{4})

Solution steps

y = (x + 1) (x + 2)

y = (x + 1) (x + 2)

The parabola parameters are:

a = 1, m = - 1, n = - 2

x_{v} = \frac{m + n}{2}

x_{v} = \frac{( - 1 ) + ( - 2 )}{2}

Simplify

x_{v} = - \frac{3}{2}

Plug in

x_{v} = - \frac{3}{2}

to find the

y_{v}

value

y_{v} = - \frac{1}{4}

Therefore the parabola vertex is

(- \frac{3}{2}, - \frac{1}{4})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- \frac{3}{2}, - \frac{1}{4})

f (x) = x^{2} + 2

f (x) = x^{2} - x

x^{2} - 8 x + 12

X^{2} - 5 X - 24

y = x^{2} + 8 x + 7

What is the vertices y=(x+1)(x+2) ?
The vertices y=(x+1)(x+2) is Minimum (-3/2 ,-1/4)